- In Webwork You Will Get Plenty Of Practice Finding The Reduced Row Echelon Form R Of A Matrix A But Here They Are Give 1 (77.77 KiB) Viewed 36 times
In webwork you will get plenty of practice finding the reduced row echelon form R of a matrix A. But here, they are given to you so we can focus on understanding how to use this information: A=⎝⎛2−242−13−30−69−5172−13−126−24⎠⎞R=⎝⎛10001000121−1010−302⎠⎞ (a) (9 points) Using R, write out what the general solution is to the equation Axn=0 below (Recall xn is a solution in the nullspace of A ). Since there are 3 free columns, you should have 3 vectors: xn=x4()+x5()+x6() (b) (5 points) Now since R, and as a consequence A, both have pivots in every row, we know that this means A has full row rank. We also know A is a 3 by 6 matrix. 'This means which of the following three things(circle the correct choice): A. Ax=b has 1 solution for every vector b. B. Ax=b has ∞ solutions for every vector b. C. Ax=b has either no solution or 1 solution for every vector b. D. Ax=b has either no solution or ∞ solutions for every vector b. (c) (6 points) Using the fact that the angmented matrix ∣A∣b∣ reduces to the angmented matrix ∣R∣d∣ where: b=⎝⎛61218⎠⎞d=⎝⎛−3−6−8⎠⎞ Find the particular solution xp where the special solutions of the nullspace are chosen to have 0 coefficients: xp=() Using x=xp+xn, we get the complete solution for the equation Ax=b for the given A and b above. (No need to write anything for this)